First Law of Thermodynamics

1
First Law of Thermodynamics

• Conservation of Energy for Thermal Systems

2
Joule Equivalent of Heat

• James Joule showed that mechanical energy could
  be converted to heat and arrived at the
  conclusion that heat was another form of energy.
• He showed that 1 calorie of heat was equivalent
  to 4.184 J of work.
• 1 cal 4.184 J

3
Energy

• Mechanical Energy KE, PE, E
• Work is done by energy transfer.
• Heat is another form of energy.
• Need to expand the conservation of energy
  principle to accommodate thermal systems.

4
1st Law of Thermodynamics

• Consider an example system of a piston and
  cylinder with an enclosed dilute gas
  characterized by P,V,T n.

5
1st Law of Thermodynamics

• What happens to the gas if the piston is moved
  inwards?

6
1st Law of Thermodynamics

• If the container is insulated the temperature
  will rise, the atoms move faster and the pressure
  rises.
• Is there more internal energy in the gas?

7
1st Law of Thermodynamics

• External agent did work in pushing the piston
  inward.
• W Fd
• (PA)Dx
• W PDV

Dx
8
1st Law of Thermodynamics

• Work done on the gas equals the change in the
  gases internal energy,
• W DU

Dx
9
1st Law of TD

• Lets change the situation
• Keep the piston fixed at its original location.
• Place the cylinder on a hot plate.
• What happens to gas?

10
Heat flows into the gas. Atoms move faster,
internal energy increases. Q heat in Joules DU
change in internal energy in Joules. Q DU
11
1st Law of TD
F

• What if we added heat and pushed the piston in at
  the same time?

12
1st Law of TD

• Work is done on the gas, heat is added to the gas
  and the internal energy of the gas increases!
• Q W DU

F
13
1st Law of TD

• Some conventions
• For the gases perspective
• heat added is positive, heat removed is negative.
• Work done on the gas is positive, work done by
  the gas is negative.
• Temperature increase means internal energy change
  is positive.

14
1st Law of TD

• Example 25 L of gas is enclosed in a
  cylinder/piston apparatus at 2 atm of pressure
  and 300 K. If 100 kg of mass is placed on the
  piston causing the gas to compress to 20 L at
  constant pressure. This is done by allowing heat
  to flow out of the gas. What is the work done on
  the gas? What is the change in internal energy
  of the gas? How much heat flowed out of the gas?

15

• Po 202,600 Pa, Vo 0.025 m3, To 300 K, Pf
  202,600 Pa, Vf0.020 m3, Tf
• n PV/RT.
• W -PDV
• DU 3/2 nRDT
• Q W DU
• W -PDV -202,600 Pa (0.020 0.025)m3
• 1013 J energy added to the gas.
• DU 3/2 nRDT1.5(2.03)(8.31)(-60)-1518 J
• Q W DU 1013 1518 -505 J heat out